SOLUTION: With the help of permutations: Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of them which are: (a) gre

Algebra ->  Permutations -> SOLUTION: With the help of permutations: Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of them which are: (a) gre      Log On


   

Question 1150509: With the help of permutations:
Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of
them which are: (a) greater than 700; (b) odd; (c) divisible by 5.
Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

In my previous post, I asked the visitor, if the three given conditions should be satisfied simultaneously or separately


     My question is here
  https://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.1150507.html


I did not get an answer.


It looks like the person who posted it, does not understand my question.


Therefore, I will make my own assumptions.


I will assume that all three conditions a), b) and c) should be satisfied simultaneously.


Then the first digit (hundreds digit) must be 7, 8, or 9 (three possibilities);

     the last digit (ones digit) must be 5, and

the middle digit (tens digit) can be any of 8 digits from 0 to 9, excluding the digit in the first position and excluding 5.


It gives the total number of cases  3*8 = 24.


ANSWER.  24 different numbers.